Question

In `triangle XYZ` if x=1, y=2, `z=sqrt5`, how do you find the exact value of cosZ?

Answer 1

`cosZ=0`

Explanation:

As `x=1`, `y=2` and `z=sqrt5`,

it is apparent that the square on the largest side `z=sqrt5` is `5` and is equal to sum of the squares on other two sides of the triangle as `5=1^2+2^2`

Hence `z` is hypotenuse and `m/_Z=pi/2`

and `cosZ=0`

Answer 2

Applying cosine law for triangle we can write

`cosZ=(x^2+y^2-z^2)/(2xy)`

`=>cosZ=(1^2+2^2-(sqrt5)^2)/(2*1*2)=(5-5)/4=0`